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Physics Question on Oscillations

A particle is executing a simple harmonic motion. Its maximum acceleration is α\alpha and maximum velocity is β\beta Then, its time period of vibration will be

A

β2α \frac{ \beta^2}{ \alpha}

B

2πβα\frac{ 2 \pi \beta}{ \alpha}

C

β2α2 \frac{ \beta^2}{ \alpha^2}

D

αβ\frac{ \alpha}{ \beta}

Answer

2πβα\frac{ 2 \pi \beta}{ \alpha}

Explanation

Solution

lf A and ω\omega be amplitude and angular frequency of vibration, then
α=ω2A\alpha = \omega^2 A \hspace20mm ...(i)
and β=ωA\beta = \omega A \hspace20mm...(ii)
Dividing eqn. (i) by eqn. (ii), we get
αβ=ω2Aω A=ω\frac{ \alpha}{ \beta} = \frac{ \omega^2 A }{ \omega \, A} = \omega
∴\therefore Time period of vibration is
T = 2πω=2π(α/β)=2πβα\frac{ 2 \pi}{ \omega} = \frac{ 2 \pi}{ ( \alpha / \beta)} = \frac{ 2 \pi \beta}{ \alpha}